One mole of `H_(2)O` and one mole of CO are taken in 10 L vessel and heated to 725 K. at equilibrium 40% of water (by mass) reacts with CO according to the equation `H_(2)O(g)+CO(g)hArr H_(2)(g)+CO_(2)(g)`.
Calculate the equilibrium constant for the reaction.

Explanation

Initial amount of `H_(2)O=1mol=18g`. If 40% of water by mass reacts with CO, then the amount of `H_(2)O` consumed `=18xx(40)/(100)=7.2g=0.4mol`
So, the amount of `H_(2)O` that remains equilibrium `=1-0.4=0.6mol`
As per the reaction, 0.4 of `H_(2)O` reacts with equal amount of `CO(g)` and forms 0.4 mol of each of `H_(2) and CO_(2)`.
So, in the equilibrium mixture, `[H_(2)O]=(0.6)/(10)mol*L^(-1)=0.06mol*L^(-1)`
`[CO]=((1-0.4))/(10)mol*L^(-1)=0.06mol*L^(-1)`
and `[H_(2)]=[CO_(2)]=(0.4)/(10)mol*L^(-1)=0.04mol*L^(-1)`
hence, for the reaction
`K_(c)=([H_(2)][CO_(2)])/([H_(2)O][CO])=(0.04xx0.04)/(0.06xx0.06)=0.44`